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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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 Posted: Sat Jan 12, 2008 8:13 am Post subject: friday 1/11 extreme #915 |
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| Code: | . . 8 | 1 . . | . . .
. . 2 | . 6 . | 5 . .
5 . . | . 9 . | . . 2
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7 4 . | . . . | 1 . .
. 8 . | 5 . 4 | . 7 .
. . 1 | . . . | . 5 3
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6 . . | . 8 . | . . 5
. . 9 | . 3 . | 7 . .
. . . | . . 6 | 3 . . |
http://www.sudoku.org.uk/DailySudoku.asp?day=11/01/2008
norm |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sat Jan 12, 2008 8:34 am Post subject: |
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| Code: | 349 379 8 | 1 5 2 | 6 349 479
1 379 2 | 348 6 378 | 5 349 479
5 367 346 | 34 9 37 | 8 1 2
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7 4 5 | 36 2 39 | 1 #689 #689
39 8 36 | 5 1 4 | 2 7 69
29 269 1 | 68 7 89 | 4 5 3
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6 23 34 | 7 8 1 | 9 24 5
48 1 9 | 2 3 5 | 7 #468 #468
28 5 7 | 9 4 6 | 3 28 1 |
am I correct that this {6,8} UR solves the puzzle?? by taking out the 8's in r8c89. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Sat Jan 12, 2008 2:23 pm Post subject: |
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I took a diversion on the two <8> pincer cells of the idle [28-29-89] xy-wing@. Both or either of the pincer cells (A,B) must contain digit <8>.
When R9C1=8 => R8C1=4 => R7C3=3 => R5C3=6(c)
Or R6C6=8 => R6C3=6(d)
Now I have two new <6> pincers in R5C3(c) and in R6C4(d), which eliminate <6> in R6C2, that solves the puzzle.
| Code: |
+--------------------------+--------------------------+--------------------------+
| 349 379 8 | 1 5 2 | 6 349 479 |
| 1 379 2 | 348 6 378 | 5 349 479 |
| 5 367 346 | 34 9 37 | 8 1 2 |
+--------------------------+--------------------------+--------------------------+
| 7 4 5 | 36 2 39 | 1 689 689 |
| 39 8 36 c | 5 1 4 | 2 7 69 |
|@29 2-[6]9 1 | 68 d 7 @89 B | 4 5 3 |
+--------------------------+--------------------------+--------------------------+
| 6 23 34 | 7 8 1 | 9 24 5 |
| 48 1 9 | 2 3 5 | 7 468 468 |
|@28 A 5 7 | 9 4 6 | 3 28 1 |
+--------------------------+--------------------------+--------------------------+ |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Sat Jan 12, 2008 3:06 pm Post subject: |
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| storm_norm wrote: | | am I correct that this {6,8} UR solves the puzzle?? by taking out the 8's in r8c89. |
There are 2 strong links for both 6 and 8 in the UR. From those i see, that r4c9<>6 and r8c8<>8.
Note, that 6 or 8 must be elsewhere in the rows/boxes/columns to avoid the DP.
From the 2 rows we can see, that either r4c4=6 (=> r6c4=8 => r6c6=9) or r8c1=8 (=> r8c6=9), so we can eliminate 9 from r6c2.
This opens an xy-wing to solve the puzzle.
Similarly in the columns or boxes either r5c9=6 or r9c8=8, but i dont see another elimnation from that.
Maybe i should mention, that also 49-689-468 in rows 2, 4 and 6 form a deadly pattern. So either r2c8=3 (=> r1c8=9) or r2c9=4 (=> r1c289=379), so 9 could be eliminated from r1c1 also this way. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sat Jan 12, 2008 3:57 pm Post subject: |
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| ravel wrote: | | storm_norm wrote: | | am I correct that this {6,8} UR solves the puzzle?? by taking out the 8's in r8c89. |
There are 2 strong links for both 6 and 8 in the UR. From those i see, that r4c9<>6 and r8c8<8> r6c4=8 => r6c6=9) or r8c1=8 (=> r8c6=9), so we can eliminate 9 from r6c2.
This opens an xy-wing to solve the puzzle.
Similarly in the columns or boxes either r5c9=6 or r9c8=8, but i dont see another elimnation from that.
Maybe i should mention, that also 49-689-468 in rows 2, 4 and 6 form a deadly pattern. So either r2c8=3 (=> r1c8=9) or r2c9=4 (=> r1c289=379), so 9 could be eliminated from r1c1 also this way. |
thank you, Ravel.
note to self...  ...put the types of UR's to memorization. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sat Jan 12, 2008 6:49 pm Post subject: |
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Some extended Medusa coloring (what else?) revealed the following AIC that solves the puzzle:
(2)r6c1-(2)r9c1=(2-3)r7c2=(3)r7c3-(3)r5c3=(3-9)r5c1=(9-2)r6c1; r6c1<>2
In ordinary language: If r6c1 is <2>, then r7c2 is <2>, r7c3 is <3>, r5c1 is <3>, and r6c1 is <9>, contradicting the assumption that r6c1 is <2>. So, r6c1 cannot be <2>. |
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